Let `g (x )=f ( x- sqrt( 1-x ^(2))) and f ' (x) =1-x ^(2) ` then `g'(x)` equal to:

Options

A. `1-x ^(2)`
B. `sqrt(1-x ^(2))`
C. `2x (x+ sqrt(1-x ^(2)))`
D. `2x (x- sqrt(1- x^(2)))`

Answer

To find \( g'(x) \) where \( g(x) = f(x - \sqrt{1 - x^2}) \) and \( f'(x) = 1 - x^2 \), we will use the chain rule for differentiation. Here’s the step-by-step solution:

### Step 1: Identify the function and its derivative
We have:
- \( g(x) = f(x - \sqrt{1 - x^2}) \)
- \( f'(x) = 1 - x^2 \)

### Step 2: Differentiate \( g(x) \) using the chain rule
Using the chain rule, we differentiate \( g(x) \):
\[ g'(x) = f'(u) \cdot \frac{du}{dx} \]
where \( u = x - \sqrt{1 - x^2} \).

### Step 3: Calculate \( \frac{du}{dx} \)
Now, we need to find \( \frac{du}{dx} \):
\[ u = x - \sqrt{1 - x^2} \]
Differentiating \( u \):
\[ \frac{du}{dx} = 1 - \frac{d}{dx}(\sqrt{1 - x^2}) \]
Using the chain rule for \( \sqrt{1 - x^2} \):
\[ \frac{d}{dx}(\sqrt{1 - x^2}) = \frac{1}{2\sqrt{1 - x^2}} \cdot (-2x) = -\frac{x}{\sqrt{1 - x^2}} \]
Thus,
\[ \frac{du}{dx} = 1 + \frac{x}{\sqrt{1 - x^2}} \]

### Step 4: Substitute back into \( g'(x) \)
Now, substituting \( u \) and \( \frac{du}{dx} \) back into the expression for \( g'(x) \):
\[ g'(x) = f'(u) \cdot \left( 1 + \frac{x}{\sqrt{1 - x^2}} \right) \]
Substituting \( u = x - \sqrt{1 - x^2} \):
\[ g'(x) = f'(x - \sqrt{1 - x^2}) \cdot \left( 1 + \frac{x}{\sqrt{1 - x^2}} \right) \]

### Step 5: Substitute \( f'(u) \)
Now, we need to evaluate \( f'(u) \):
\[ f'(u) = 1 - (x - \sqrt{1 - x^2})^2 \]
Calculating \( (x - \sqrt{1 - x^2})^2 \):
\[ (x - \sqrt{1 - x^2})^2 = x^2 - 2x\sqrt{1 - x^2} + (1 - x^2) = 1 - 2x\sqrt{1 - x^2} \]
Thus,
\[ f'(u) = 1 - (1 - 2x\sqrt{1 - x^2}) = 2x\sqrt{1 - x^2} \]

### Step 6: Final expression for \( g'(x) \)
Now substituting \( f'(u) \) into the equation for \( g'(x) \):
\[ g'(x) = 2x\sqrt{1 - x^2} \cdot \left( 1 + \frac{x}{\sqrt{1 - x^2}} \right) \]
Distributing:
\[ g'(x) = 2x\sqrt{1 - x^2} + 2x^2 \]

### Final Result
Thus, the derivative \( g'(x) \) is:
\[ g'(x) = 2x\sqrt{1 - x^2} + 2x^2 \]